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MatroZZZ [7]
3 years ago
9

List the following minerals in order of decreasing hardness: apatite, calcite, corundum, feldspar, and talc.

Chemistry
2 answers:
ahrayia [7]3 years ago
7 0

Explanation:

Mohs scale is a scale that represents hardness of different minerals in increasing order from bottom to top.

So, according to Mohs scale hardness of  given minerals is rated as follows.

Apaptite is at number 5.

Calcite is at number 3.

Corundum is at number 9.

Feldspar is at number 6.

Talc is at number 1.

Therefore, given minerals in order of decreasing hardness are listed as follows.

        Corundum > Feldspar > Apaptite > Calcite > Talc

Leto [7]3 years ago
5 0
Corundum, Feldspar, Apatite, Calcite, and Talc.
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A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mix
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Answer:

M_f=38.8\%

Explanation:

From the question we are told that:

Pressure P=747mmHg

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Volume V=11.1

Heat Produced Q=780kJ

Generally the equation for ideal gas is mathematically given by

 PV=nRT

 n= (747/760) *11.1/ (0.0821*298)

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Therefore

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 x=0.446-y .....1

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

 x(889) + y(2220) = 760 ...... 2

Comparing Equation 1 and 2 and solving simultaneously

 x=0.446-y .....1

 x(889) + y(2220) = 760 ...... 2

 x=0.173

 y=0.273

Therefore

Mole fraction 0f Methane is mathematically given as

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1.33 dm3 of water at 70°C are saturated by 2.25
astraxan [27]

Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C  = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

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